Projective Geometry - Key Proof

ProofProof of Desargues' Theorem in Projective 3-Space

Theorem: In projective 3-space, if triangles $\triangle ABC$ and $\triangle A'B'C'$ are perspective from a point $O$ (lines $AA'$ , $BB'$ , $CC'$ concurrent at $O$ ), then they are perspective from a line (points $P = AB \cap A'B'$ , $Q = BC \cap B'C'$ , $R = CA \cap C'A'$ are collinear).

Proof:

Assume the two triangles lie in distinct planes $\pi$ and $\pi'$ (if they're coplanar, the theorem requires separate proof or becomes an axiom).

Step 1 (Configuration): Since $AA'$ , $BB'$ , $CC'$ all pass through $O$ , and $A, B, C \in \pi$ while $A', B', C' \in \pi'$ , the point $O$ serves as the "center of perspectivity."

Step 2 (Finding intersection points):

Point $P$ lies on line $AB$ (in plane $\pi$ ) and on line $A'B'$ (in plane $\pi'$ )
Therefore, $P$ lies in both planes: $P \in \pi \cap \pi'$
Similarly, $Q \in \pi \cap \pi'$ and $R \in \pi \cap \pi'$

Step 3 (Collinearity): The intersection of two distinct planes in projective 3-space is a line. Since $P$ , $Q$ , $R$ all lie in $\pi \cap \pi'$ , they are collinear. ∎

This proof showcases the power of working in higher dimensions. The theorem becomes almost trivial in 3-space, reducing to the elementary fact that two planes intersect in a line.

■

The elegance of this proof demonstrates why Desargues' theorem holds automatically in projective 3-space and higher dimensions. The challenge is proving it in the projective plane itself, where it becomes an independent axiom or requires algebraic methods.

Remark

Proof in the Projective Plane (requires different approach):

When both triangles lie in the same plane, the 3D argument doesn't apply. Instead, one must either:

Assume Desargues as an axiom (synthetic approach)
Use coordinates and verify algebraically
Embed in 3-space by lifting points appropriately

The coordinateization approach shows that Desargues holds if and only if the underlying algebraic structure is a division ring (with associativity).

Converse Proof: To prove the converse (perspective from a line implies perspective from a point), we use duality or a similar construction. If $P$ , $Q$ , $R$ are collinear on line $\ell$ , consider the planes determined by:

Triangle $ABC$ determines plane $\pi$
Triangle $A'B'C'$ determines plane $\pi'$
Line $\ell$ lies in both planes (since $P, Q, R$ are in both)

The lines $AA'$ , $BB'$ , $CC'$ must meet at a point (the intersection of certain planes constructed from the given data).

ExampleCoordinate Verification

Using homogeneous coordinates in $\mathbb{P}^2(\mathbb{R})$ , let:

\begin{align} A &= [1:0:0], \quad B = [0:1:0], \quad C = [0:0:1] \\ O &= [1:1:1] \end{align}

For lines $AA'$ , $BB'$ , $CC'$ to pass through $O$ , we can parametrize:

A' = [1:a:a], \quad B' = [b:1:b], \quad C' = [c:c:1]

Computing intersections $P$ , $Q$ , $R$ and verifying collinearity involves determinant calculations showing that the three points lie on a common line. This confirms the theorem algebraically.

The synthetic proof without coordinates uses auxiliary constructions and repeated applications of basic incidence axioms. Such proofs, while longer, reveal the purely geometric content without relying on algebraic tools. Hilbert's Grundlagen der Geometrie contains elegant synthetic proofs of classical theorems including Desargues.